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Question Number 95215 by mathmax by abdo last updated on 24/May/20
calculate Σ_(n=0) ^∞  (((−1)^n )/(4n+1))
$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}} \\ $$
Answered by mathmax by abdo last updated on 25/May/20
let s(x) =Σ_(n=0) ^∞  (((−1)^n )/(4n+1))x^(4n+1)   with ∣x∣ ≤1 and x≠−1  s^′ (x) =Σ_(n=0) ^∞  (−1)^n  x^(4n)  =Σ_(n=0) ^∞ (−x^4 )^n  =(1/(1+x^4 )) ⇒  s(x) =∫_0 ^x  (dt/(t^4  +1)) +c    s(0) =0 =c ⇒s(x) =∫_0 ^x  (dt/(t^4  +1)) and  Σ_(n=0) ^∞  (((−1)^n )/(4n+1)) =∫_0 ^1    (dt/(t^4  +1)) =∫_0 ^1    ((1/t^2 )/(t^2  +(1/t^2 )))dt  =(1/2)∫_0 ^1  ((1+(1/t^2 )−1+(1/t^2 ))/(t^2  +(1/t^2 )))dt =(1/2)∫_0 ^1  ((1+(1/t^2 ))/(t^2  +(1/t^2 )))dt−(1/2)∫_0 ^1  ((1−(1/t^2 ))/(t^2  +(1/t^2 )))dt  =(1/2) ∫_0 ^1  ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt(→t−(1/t)=u)−(1/2) ∫_0 ^1   ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt(→t+(1/t)=v)  =(1/2) ∫_(−∞) ^0  (du/(u^2  +2)) −(1/2) ∫_(+∞) ^2  (dv/(v^2 −2))  we have  ∫_(−∞) ^0  (du/(u^2  +2)) =_(u =(√2)α)    ∫_(−∞) ^0  (((√2)dα)/(2(1+α^2 ))) =(1/( (√2))) [arctanα]_(−∞) ^0   =(1/( (√2)))((π/2)) =(π/(2(√2))) and  ∫_∞ ^2  (dv/(v^2 −2)) =−∫_2 ^∞  (dv/((v−(√2))(v+(√2))))  =−(1/(2(√2)))∫_2 ^(+∞) ((1/(v−(√2)))−(1/(v+(√2))))dv =−(1/(2(√2)))[ln∣((v−(√2))/(v+(√2)))∣]_2 ^(+∞)   =−(1/(2(√2)))(−ln(((2−(√2))/(2(√2))))) =(1/(2(√2)))ln((1/( (√2)))−(1/2)) ⇒  Σ_(n=0) ^∞  (((−1)^n )/(4n+1)) =(π/(4(√2))) −(1/(4(√2)))ln((1/( (√2)))−(1/2))
$$\mathrm{let}\:\mathrm{s}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}}\mathrm{x}^{\mathrm{4n}+\mathrm{1}} \:\:\mathrm{with}\:\mid\mathrm{x}\mid\:\leqslant\mathrm{1}\:\mathrm{and}\:\mathrm{x}\neq−\mathrm{1} \\ $$$$\mathrm{s}^{'} \left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{4n}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{s}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:+\mathrm{c}\:\:\:\:\mathrm{s}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\mathrm{c}\:\Rightarrow\mathrm{s}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:\mathrm{and} \\ $$$$\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}}\mathrm{dt}\left(\rightarrow\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}=\mathrm{u}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} −\mathrm{2}}\mathrm{dt}\left(\rightarrow\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}=\mathrm{v}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{\mathrm{0}} \:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{+\infty} ^{\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{2}}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{−\infty} ^{\mathrm{0}} \:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{2}}\:=_{\mathrm{u}\:=\sqrt{\mathrm{2}}\alpha} \:\:\:\int_{−\infty} ^{\mathrm{0}} \:\frac{\sqrt{\mathrm{2}}\mathrm{d}\alpha}{\mathrm{2}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\mathrm{arctan}\alpha\right]_{−\infty} ^{\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{and}\:\:\int_{\infty} ^{\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{2}}\:=−\int_{\mathrm{2}} ^{\infty} \:\frac{\mathrm{dv}}{\left(\mathrm{v}−\sqrt{\mathrm{2}}\right)\left(\mathrm{v}+\sqrt{\mathrm{2}}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{2}} ^{+\infty} \left(\frac{\mathrm{1}}{\mathrm{v}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{v}+\sqrt{\mathrm{2}}}\right)\mathrm{dv}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{ln}\mid\frac{\mathrm{v}−\sqrt{\mathrm{2}}}{\mathrm{v}+\sqrt{\mathrm{2}}}\mid\right]_{\mathrm{2}} ^{+\infty} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(−\mathrm{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}}\:=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$

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